Plane Geometry Practice Questions — ACT — Page 9

Question 161Question →

In isosceles trapezoid $ABCD$ , the shorter base $AB$ measures $7$ units and the longer base $CD$ measures $17$ units. The congruent legs $AD$ and $BC$ each form a $45^\circ$ angle with base $CD$ . What is the length of diagonal $AC$ ?

13

17

5\sqrt{2}

\sqrt{119}

169

Show answer & explanation

Answer: $13$

Answer

13 units

Dropping altitude

AP

perpendicular to base

CD

divides base

CD

into

DP = 5

units and

PC = 12

units. Since triangle

APD

is a

45^\circ-45^\circ-90^\circ

right triangle, height

AP = DP = 5

. Right triangle

APC

has legs

AP = 5

and

PC = 12

, making hypotenuse

AC = \sqrt{5^2 + 12^2} = 13

.

Step-by-Step Solution

1

Find the length of the base projection segment for the isosceles trapezoid.

Segment

DP = 5

units.

Draw altitude

AP

perpendicular to

CD

. Because trapezoid

ABCD

is isosceles, the projection

DP = \frac{CD - AB}{2} = \frac{17 - 7}{2} = 5

.

2

Determine the altitude of the trapezoid using special right triangle properties.

Altitude

AP = 5

units.

Triangle

APD

is a

45^\circ-45^\circ-90^\circ

right triangle, so its legs are congruent (

AP = DP = 5

).

3

Calculate the length of the remaining base segment in right triangle

APC

.

Segment

PC = 12

units.

Segment

PC = CD - DP = 17 - 5 = 12

.

4

Apply the Pythagorean Theorem to right triangle

APC

to solve for diagonal

AC

.

Diagonal

AC = 13

units.

AC = \sqrt{AP^2 + PC^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13

.

Key Concept

Pythagorean Theorem and Special Right Triangles

Estimated Time:1m 15s

Question 162Question →

In the standard $(x, y)$ coordinate plane, trapezoid $ABCD$ has vertices at $A(0, 0)$ , $B(4, 8)$ , $C(12, 8)$ , and $D(16, 0)$ . Point $P$ is the midpoint of diagonal $AC$ , and point $Q$ is the midpoint of diagonal $BD$ . What is the distance, in coordinate units, between point $P$ and point $Q$ ?

2

4

6

8

12

Show answer & explanation

Answer: $4$

Answer

The distance between point

P

and point

Q

is

4

units.

The correct answer is

4

. Using the midpoint formula, the midpoint of diagonal

AC

is

P(6, 4)

and the midpoint of diagonal

BD

is

Q(10, 4)

. Because both points lie on the horizontal line

y = 4

, the distance between them is

|10 - 6| = 4

units.

Step-by-Step Solution

1

Calculate the coordinates of midpoint

P

of diagonal

AC

P = \left(\frac{0 + 12}{2}, \frac{0 + 8}{2}\right) = (6, 4)

The midpoint formula is

\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)

.

2

Calculate the coordinates of midpoint

Q

of diagonal

BD

Q = \left(\frac{4 + 16}{2}, \frac{8 + 0}{2}\right) = (10, 4)

Applying the midpoint formula to vertices

B(4, 8)

and

D(16, 0)

.

3

Calculate the horizontal distance between

P(6, 4)

and

Q(10, 4)

\text{Distance } PQ = \sqrt{(10 - 6)^2 + (4 - 4)^2} = 10 - 6 = 4

Since both midpoints share the same

y

-coordinate (

y = 4

), the distance is simply the absolute difference between their

x

-coordinates.

Key Concept

Midpoints of Diagonals in a Trapezoid

Alternative Method

For any trapezoid with parallel bases of lengths

b_1

and

b_2

(where

b_1 > b_2

), the length of the segment connecting the midpoints of the diagonals is given by the formula

\frac{b_1 - b_2}{2}

. Here

b_1 = 16 - 0 = 16

and

b_2 = 12 - 4 = 8

, so the length is

\frac{16 - 8}{2} = 4

.

Estimated Time:1m 15s

Question 163Question →

In a circle with center $O$ , sector $AOB$ has an area of $18\pi$ square centimeters and an arc length along $\widehat{AB}$ of $3\pi$ centimeters. A straight line segment $AB$ is drawn to complete triangle $AOB$ . What is the area, in square centimeters, of the circular segment bounded by line segment $AB$ and arc $\widehat{AB}$ ?

18\pi - 36\sqrt{2}

18\pi - 72\sqrt{2}

18\pi - 36

36\pi - 18\sqrt{2}

18\pi - 72

Show answer & explanation

Answer: $18\pi - 36\sqrt{2}$

Answer

The area of the circular segment is

18\pi - 36\sqrt{2}

square centimeters.

The expression

18\pi - 36\sqrt{2}

correctly represents the area of the circular segment. By dividing sector area (

18\pi

) by arc length (

3\pi

), we obtain

\frac{1}{2}r = 6

, giving a radius

r = 12\text{ cm}

. Substituting

r = 12

into

r\theta = 3\pi

gives central angle

\theta = \frac{\pi}{4}

radians (

45^\circ

). The area of triangle

AOB

is

\frac{1}{2}(12)^2\sin(45^\circ) = 36\sqrt{2}

. Subtracting the triangle area from the sector area yields

18\pi - 36\sqrt{2}

.

Step-by-Step Solution

1

Relate sector area and arc length formulas to find radius

r

Sector area

A = \frac{1}{2}r^2\theta = 18\pi

and arc length

s = r\theta = 3\pi

. Dividing sector area by arc length gives

\frac{\frac{1}{2}r^2\theta}{r\theta} = \frac{18\pi}{3\pi} \implies \frac{1}{2}r = 6 \implies r = 12\text{ cm}

.

Dividing the sector area equation by the arc length equation isolates the radius

r

.

2

Find central angle

\theta

s = r\theta \implies 3\pi = 12\theta \implies \theta = \frac{3\pi}{12} = \frac{\pi}{4}\text{ radians}

(

45^\circ

).

Knowing the radius

r

allows calculating

\theta

directly from the arc length formula.

3

Calculate the area of triangle

AOB

\text{Area}(AOB) = \frac{1}{2}r^2\sin\theta = \frac{1}{2}(12)^2\sin\left(\frac{\pi}{4}\right) = 72 \cdot \frac{\sqrt{2}}{2} = 36\sqrt{2}\text{ cm}^2.

The area of a triangle with two sides of length

r

and included angle

\theta

is

\frac{1}{2}r^2\sin\theta

.

4

Subtract the triangle area from the sector area to find the segment area

\text{Segment Area} = \text{Area}(\text{sector } AOB) - \text{Area}(\triangle AOB) = 18\pi - 36\sqrt{2}\text{ cm}^2.

The region bounded by the chord and the arc is the sector minus the central triangle.

Key Concept

The area of a circular segment is found by subtracting the area of the central triangle (

\frac{1}{2}r^2\sin\theta

) from the area of the circular sector (

\frac{1}{2}r^2\theta

).

Question 164Question →

In kite $ABCD$ , diagonals $AC$ and $BD$ intersect perpendicularly at point $P$ . If $AP = 9$ centimeters, $PC = 16$ centimeters, and $BP = PD = 12$ centimeters, what is the perimeter, in centimeters, of kite $ABCD$ ?

Show answer & explanation

Answer: 70

Answer

The perimeter of kite

ABCD

is 70 centimeters.

The diagonals of a kite intersect at right angles (

90^\circ

). Applying the Pythagorean theorem to right triangle

APB

with legs

9

cm and

12

cm gives hypotenuse

AB = 15

cm. Applying the Pythagorean theorem to right triangle

BPC

with legs

16

cm and

12

cm gives hypotenuse

BC = 20

cm. Since a kite has two pairs of equal adjacent sides (

AB = AD = 15

cm and

BC = CD = 20

cm), the total perimeter is

15 + 15 + 20 + 20 = 70

cm.

Step-by-Step Solution

1

Identify right triangles formed by the perpendicular diagonals

Four right triangles are formed:

\triangle APB

,

\triangle BPC

,

\triangle CPD

, and

\triangle DPA

.

Diagonals of a kite are perpendicular to each other.

2

Calculate upper side length

AB

AB = \sqrt{9^2 + 12^2} = \sqrt{225} = 15

cm

Apply the Pythagorean theorem

a^2 + b^2 = c^2

to right triangle

APB

.

3

Calculate lower side length

BC

BC = \sqrt{16^2 + 12^2} = \sqrt{400} = 20

cm

Apply the Pythagorean theorem

a^2 + b^2 = c^2

to right triangle

BPC

.

4

Compute the total perimeter

Perimeter

= 2(15) + 2(20) = 30 + 40 = 70

cm

A kite has two pairs of congruent adjacent sides (

AD = AB

and

CD = BC

).

Key Concept

Perpendicular diagonals and side length properties of a kite

Alternative Method

Instead of calculating all four sides individually, calculate one side from each distinct right triangle (

15

cm and

20

cm) and multiply their sum by

2

, using the property that a kite has two symmetric pairs of congruent adjacent sides:

2 \times (15 + 20) = 70

cm.

Estimated Time:1m 15s

Question 165Question →

In isosceles trapezoid $ABCD$ , side $AB$ is parallel to side $CD$ . If the measure of $\angle A$ is $(3x + 10)^\circ$ and the measure of $\angle C$ is $(5x - 30)^\circ$ , what is the measure of $\angle B$ ?

70^\circ

75^\circ

85^\circ

95^\circ

110^\circ

Show answer & explanation

Answer: $85^\circ$

Answer

The measure of

\angle B

is

85^\circ

.

The answer of

85^\circ

is correct because parallel sides

AB

and

CD

imply that consecutive interior angles

\angle A

and

\angle C

add up to

180^\circ

. Solving

(3x + 10) + (5x - 30) = 180

yields

x = 25

. Substituting

x = 25

into the expression for

\angle A

gives

3(25) + 10 = 85^\circ

. Because

ABCD

is an isosceles trapezoid, the base angles

\angle A

and

\angle B

adjacent to base

AB

are congruent, so

\angle B = 85^\circ

.

Step-by-Step Solution

1

Identify the relationship between

\angle A

and

\angle C

Since

AB \parallel CD

, angles

\angle A

and

\angle C

are consecutive interior angles along transversal

AC

(or leg

AD

), which means they are supplementary:

\angle A + \angle C = 180^\circ

.

Parallel lines cut by a transversal form supplementary consecutive interior angles.

2

Set up and solve the algebraic equation for

x

(3x + 10) + (5x - 30) = 180 \implies 8x - 20 = 180 \implies 8x = 200 \implies x = 25

.

Combine like terms and solve for

x

.

3

Calculate the measure of

\angle A

\angle A = 3(25) + 10 = 75 + 10 = 85^\circ

.

Substitute

x = 25

back into the expression for

\angle A

.

4

Determine the measure of

\angle B

using isosceles trapezoid properties

\angle B = \angle A = 85^\circ

.

In an isosceles trapezoid with

AB \parallel CD

, base angles along the same parallel base are congruent.

Key Concept

Properties of Isosceles Trapezoids and Consecutive Interior Angles

Question 166Question →

In quadrilateral $ABCD$ , $\angle B = 90^\circ$ and $\angle D = 90^\circ$ . If $AB = 12$ units, $BC = 16$ units, and $AD = 10$ units, what is the length, in units, of side $CD$ ?

10\sqrt{3}

10\sqrt{5}

6\sqrt{19}

18

10

Show answer & explanation

Answer: $10\sqrt{3}$

Answer

10\sqrt{3}

units

Connecting vertex

A

to vertex

C

creates two right-angled triangles sharing hypotenuse

AC

. In right triangle

ABC

, the Pythagorean theorem yields

AC^2 = 12^2 + 16^2 = 400

, so

AC = 20

. Next, in right triangle

ADC

,

AC

serves as the hypotenuse and

AD = 10

is one leg. Solving for leg

CD

gives

CD^2 = AC^2 - AD^2 = 20^2 - 10^2 = 300

, which simplifies to

CD = \sqrt{300} = 10\sqrt{3}

.

Step-by-Step Solution

1

Draw diagonal

AC

to divide quadrilateral

ABCD

into two right triangles,

\triangle ABC

and

\triangle ADC

, sharing hypotenuse

AC

.

Two right-angled triangles

\triangle ABC

(with right angle at

B

) and

\triangle ADC

(with right angle at

D

).

Diagonal

AC

acts as the hypotenuse for both right triangles.

2

Apply the Pythagorean Theorem to right triangle

\triangle ABC

to calculate the length of hypotenuse

AC

.

AC = \sqrt{AB^2 + BC^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20

units.

In right triangle

\triangle ABC

,

AB

and

BC

are legs.

3

Apply the Pythagorean Theorem to right triangle

\triangle ADC

to solve for leg

CD

.

CD = \sqrt{AC^2 - AD^2} = \sqrt{20^2 - 10^2} = \sqrt{400 - 100} = \sqrt{300} = 10\sqrt{3}

units.

In right triangle

\triangle ADC

,

AC

is the hypotenuse (

20

) and

AD

is a leg (

10

).

Key Concept

Multi-step applications of the Pythagorean Theorem using shared boundary hypotenuses.

Estimated Time:1m 15s

Question 167Question →

A region $ABCD$ is bounded by two concentric circular arcs of radii $R$ and $r$ (where $R > r$ ) and two radial line segments, all sharing a central angle of $72^\circ$ . The area of region $ABCD$ is $60\pi\text{ cm}^2$ , and the total perimeter of region $ABCD$ is $(12\pi + 20)\text{ cm}$ . What is the value of the outer radius $R$ , in centimeters?

10

15

20

25

30

Show answer & explanation

Answer: 20

Answer

The outer radius

R

is 20 centimeters.

The central angle of

72^\circ

represents

\frac{72}{360} = \frac{1}{5}

of a circle. The area of the region is

\frac{\pi}{5}(R^2 - r^2) = 60\pi

, which simplifies to

R^2 - r^2 = 300

, or

(R+r)(R-r) = 300

. The total perimeter is the sum of the outer arc

\frac{2\pi R}{5}

, the inner arc

\frac{2\pi r}{5}

, and the two straight side segments

2(R-r)

. Setting

\frac{2\pi}{5}(R+r) + 2(R-r) = 12\pi + 20

yields

\frac{2}{5}(R+r) = 12 \implies R+r = 30

and

2(R-r) = 20 \implies R-r = 10

. Solving the system

R+r = 30

and

R-r = 10

by adding the equations gives

2R = 40

, so

R = 20\text{ cm}

.

Step-by-Step Solution

1

Determine the fraction of the full circle represented by the

72^\circ

central angle.

The fraction is

\frac{72^\circ}{360^\circ} = \frac{1}{5}

.

Arc lengths and sector areas are proportional to the ratio of the central angle to

360^\circ

.

2

Set up and simplify the equation for the area of the region

ABCD

.

\frac{1}{5}\pi R^2 - \frac{1}{5}\pi r^2 = 60\pi \implies R^2 - r^2 = 300 \implies (R+r)(R-r) = 300

.

The region's area is the difference between the outer sector area and inner sector area.

3

Set up and simplify the equation for the perimeter of region

ABCD

.

\frac{1}{5}(2\pi R) + \frac{1}{5}(2\pi r) + 2(R - r) = 12\pi + 20 \implies \frac{2\pi}{5}(R + r) + 2(R - r) = 12\pi + 20

.

The perimeter consists of the outer arc, the inner arc, and two radial segments each of length

R - r

.

4

Equate corresponding rational and

\pi

-coefficient terms to solve for

(R+r)

and

(R-r)

.

\frac{2}{5}(R + r) = 12 \implies R + r = 30

, and

2(R - r) = 20 \implies R - r = 10

.

Equating the algebraic components yields a system of linear equations.

5

Solve the system of equations for the outer radius

R

.

Adding the equations gives

(R + r) + (R - r) = 30 + 10 \implies 2R = 40 \implies R = 20\text{ cm}

.

Eliminating

r

isolates the required variable

R

.

Key Concept

Arc length and sector area of concentric circular regions

Estimated Time:2m 0s

Question 168Question →

In right triangle $ABC$ , the right angle is located at vertex $B$ . Point $D$ lies on leg $BC$ such that $AB = 12$ inches and $BD = 9$ inches. If segment $AD$ is equal in length to segment $DC$ , what is the length of leg $BC$ , in inches?

Show answer & explanation

Answer: 24

Answer

The length of leg

BC

is 24 inches.

Applying the Pythagorean theorem to right triangle

ABD

gives

AD = \sqrt{12^2 + 9^2} = 15

inches. Because segment

AD

equals segment

DC

,

DC

is also 15 inches. Adding the lengths of segments

BD

and

DC

gives

BC = 9 + 15 = 24

inches.

Step-by-Step Solution

1

Calculate the length of hypotenuse

AD

in right triangle

ABD

AD = 15

inches

Apply the Pythagorean theorem:

AD = \sqrt{AB^2 + BD^2} = \sqrt{12^2 + 9^2} = 15

.

2

Determine the length of segment

DC

DC = 15

inches

It is given that segment

AD

is equal in length to segment

DC

.

3

Calculate the total length of leg

BC

BC = 24

inches

Add the adjacent segment lengths along leg

BC

:

BC = BD + DC = 9 + 15 = 24

.

Key Concept

Applying the Pythagorean theorem to adjacent right triangles within geometric figures.

Question 169Question →

Determine whether the following statement is true or false:

The quadrilateral formed by connecting the midpoints of the four consecutive sides of any rhombus is always a rectangle.

True

False

Show answer & explanation

Answer: True

Answer

The statement is true because the diagonals of a rhombus are perpendicular, which forces the adjacent sides of the midpoint quadrilateral (which are parallel to the diagonals) to meet at right angles.

The statement is true because the diagonals of any rhombus are perpendicular. By the Midsegment Theorem, the sides of the quadrilateral formed by connecting adjacent midpoints are parallel to these diagonals, ensuring all four interior angles are

90^\circ

, which satisfies the definition of a rectangle.

Step-by-Step Solution

1

Apply the Triangle Midsegment Theorem to rhombus

ABCD

with diagonals

AC

and

BD

.

Let

P, Q, R,

and

S

be the midpoints of sides

AB, BC, CD,

and

DA

respectively. Segment

PQ

is parallel to diagonal

AC

and has length

\frac{1}{2}AC

. Segment

QR

is parallel to diagonal

BD

and has length

\frac{1}{2}BD

.

A line segment connecting the midpoints of two sides of a triangle is parallel to the third side and half its length.

2

Use the geometric properties of a rhombus's diagonals.

In any rhombus, the diagonals

AC

and

BD

are perpendicular to each other (

AC \perp BD

).

Perpendicular diagonals are a key defining property of all rhombuses.

3

Determine the interior angle measures of quadrilateral

PQRS

.

Because

PQ \parallel AC

and

QR \parallel BD

, and

AC \perp BD

, the sides

PQ

and

QR

must be perpendicular (

PQ \perp QR

). Therefore, all four interior angles of quadrilateral

PQRS

are

90^\circ

.

If two lines are parallel to two mutually perpendicular lines, they are also mutually perpendicular.

4

Classify quadrilateral

PQRS

.

Quadrilateral

PQRS

has four right angles, so it is by definition a rectangle.

Any quadrilateral with four right angles is a rectangle.

Key Concept

Properties of Rhombus Diagonals and Midpoint Quadrilaterals

Estimated Time:1m 0s

Question 170Question →

A park planner is designing a triangular walking path $ABC$ where the corner at vertex $B$ forms a $90^\circ$ angle. A straight path $AD$ is constructed from vertex $A$ to point $D$ on side $BC$ , dividing angle $\angle BAC$ into two equal angles measuring $30^\circ$ each. If side $AB = 18$ meters, what is the length, in meters, of segment $DC$ ?

6\sqrt{3}

9\sqrt{3}

12\sqrt{3}

18\sqrt{3}

36

Show answer & explanation

Answer: $12\sqrt{3}$

Answer

The length of segment

DC

is

12\sqrt{3}

meters.

In right triangle

ABD

,

\angle BAD = 30^\circ

and

AB = 18

, so

BD = \frac{18}{\sqrt{3}} = 6\sqrt{3}

meters. In right triangle

ABC

,

\angle BAC = 60^\circ

, so

BC = 18\sqrt{3}

meters. Subtracting

BD

from

BC

gives

DC = 18\sqrt{3} - 6\sqrt{3} = 12\sqrt{3}

meters.

Step-by-Step Solution

1

Determine the angles in right triangle

ABD

and right triangle

ABC

.

In

\triangle ABD

,

\angle B = 90^\circ

and

\angle BAD = 30^\circ

. In

\triangle ABC

,

\angle B = 90^\circ

and

\angle BAC = 30^\circ + 30^\circ = 60^\circ

.

Path

AD

bisects

\angle BAC

into two

30^\circ

angles.

2

Calculate the length of segment

BD

using the

30^\circ-60^\circ-90^\circ

right triangle ratio in

\triangle ABD

.

BD = \frac{AB}{\sqrt{3}} = \frac{18}{\sqrt{3}} = 6\sqrt{3}

meters.

In a

30^\circ-60^\circ-90^\circ

triangle, the leg opposite the

30^\circ

angle is equal to the adjacent leg divided by

\sqrt{3}

.

3

Calculate the total length of leg

BC

using the

30^\circ-60^\circ-90^\circ

right triangle ratio in

\triangle ABC

.

BC = AB \cdot \sqrt{3} = 18\sqrt{3}

meters.

In

\triangle ABC

, the leg opposite the

60^\circ

angle (

BC

) is

\sqrt{3}

times the adjacent leg (

AB = 18

).

4

Subtract segment

BD

from total leg

BC

to find segment

DC

.

DC = BC - BD = 18\sqrt{3} - 6\sqrt{3} = 12\sqrt{3}

meters.

Segment addition postulate state that

BD + DC = BC

.

Key Concept

Properties of

30^\circ-60^\circ-90^\circ

Special Right Triangles

Estimated Time:1m 30s

Question 171Question →

A tangent line segment $\overline{PT}$ touches a circle at point $T$ . A secant line from external point $P$ intersects the circle at points $A$ and $B$ , such that point $A$ lies on segment $\overline{PB}$ . If $m\angle P = 35^\circ$ and the measure of minor arc $AT$ is $50^\circ$ , what is the degree measure of inscribed angle $\angle TAB$ ?

Show answer & explanation

Answer: 60

Answer

The degree measure of inscribed angle

\angle TAB

is 60 degrees.

According to the exterior angle theorem for circles, the angle formed by a tangent and a secant meeting at an external point

P

is equal to half the difference of the intercepted arcs:

m\angle P = \frac{1}{2}(m\widehat{BT} - m\widehat{AT})

. Substituting

m\angle P = 35^\circ

and

m\widehat{AT} = 50^\circ

into the equation gives

35^\circ = \frac{1}{2}(m\widehat{BT} - 50^\circ)

, which simplifies to

m\widehat{BT} = 120^\circ

. The inscribed angle

\angle TAB

intercepts arc

BT

. By the inscribed angle theorem, the measure of an inscribed angle is half the measure of its intercepted arc, giving

m\angle TAB = \frac{1}{2}(120^\circ) = 60^\circ

. Alternatively, inside triangle

PAT

, the tangent-chord angle

PTA

intercepts arc

AT

, so

m\angle PTA = \frac{1}{2}(50^\circ) = 25^\circ

. Since the angles in triangle

PAT

sum to

180^\circ

,

m\angle PAT = 180^\circ - (35^\circ + 25^\circ) = 120^\circ

. Angle

TAB

is supplementary to angle

PAT

, so

m\angle TAB = 180^\circ - 120^\circ = 60^\circ

.

Step-by-Step Solution

1

Use the exterior angle relationship for the secant and tangent to find the measure of arc

BT

.

m\widehat{BT} = 120^\circ

The exterior angle measure equals half the difference of intercepted arcs

BT

and

AT

:

35^\circ = \frac{1}{2}(m\widehat{BT} - 50^\circ)

.

2

Use the Inscribed Angle Theorem to find

m\angle TAB

.

m\angle TAB = 60^\circ

An inscribed angle measure is equal to half the measure of its intercepted arc:

m\angle TAB = \frac{1}{2}(120^\circ) = 60^\circ

.

Key Concept

Secant-Tangent Angle Theorem and Inscribed Angle Theorem

Estimated Time:1m 30s

Question 172Question →

A rhombus-shaped garden plot $ABCD$ has side lengths of $15$ feet each. The length of the shorter diagonal, $AC$ , is $18$ feet. A gardener places a straight divider line along the longer diagonal, $BD$ . What is the length, in feet, of the divider line along diagonal $BD$ ?

Show answer & explanation

Answer: 24

Answer

The length of the divider line along diagonal

BD

is 24 feet.

The diagonals of a rhombus are perpendicular bisectors of each other. The point of intersection

E

creates right triangle

AEB

, where the hypotenuse is rhombus side

AB = 15

feet and one leg is

AE = \frac{18}{2} = 9

feet. Applying the Pythagorean Theorem yields

9^2 + BE^2 = 15^2

, which simplifies to

81 + BE^2 = 225

, giving

BE = 12

feet. Doubling

BE

gives the complete length of diagonal

BD = 24

feet.

Step-by-Step Solution

1

Identify geometric properties of a rhombus regarding its diagonals.

The diagonals of rhombus

ABCD

are perpendicular to each other and bisect each other at intersection point

E

.

In any rhombus, the diagonals act as perpendicular bisectors, forming four right triangles.

2

Calculate the leg length

AE

in right triangle

AEB

.

AE = \frac{18}{2} = 9

feet.

Point

E

is the midpoint of diagonal

AC

.

3

Use the Pythagorean Theorem to calculate leg length

BE

.

9^2 + BE^2 = 15^2 \implies 81 + BE^2 = 225 \implies BE^2 = 144 \implies BE = 12

feet.

In right triangle

AEB

, side

AB = 15

is the hypotenuse, and

AE = 9

is one leg.

4

Find the total length of diagonal

BD

.

BD = 2 \times BE = 2 \times 12 = 24

feet.

Point

E

bisects diagonal

BD

, so

BD

is twice the length of

BE

.

Key Concept

Using the Pythagorean Theorem on right triangles formed by the perpendicular bisecting diagonals of a rhombus.

Question 173Question →

Kite $WXYZ$ has perpendicular diagonals $WY$ and $XZ$ that intersect at point $M$ . Diagonal $WY$ bisects diagonal $XZ$ such that $XM = MZ = 8$ centimeters. If $WM = 6$ centimeters and $MY = 15$ centimeters, what is the perimeter of kite $WXYZ$ , in centimeters?

27

46

54

68

74

Show answer & explanation

Answer: 54

Answer

The perimeter of kite

WXYZ

is 54 centimeters.

The correct answer is 54. The perpendicular diagonals of a kite form four interior right triangles. Using the legs

WM = 6

cm and

XM = 8

cm, the upper side

WX

is

\sqrt{6^2 + 8^2} = 10

cm. Using the legs

MY = 15

cm and

XM = 8

cm, the lower side

YX

is

\sqrt{15^2 + 8^2} = 17

cm. Summing all four outer sides (

10 + 10 + 17 + 17

) gives a total perimeter of 54 cm.

Step-by-Step Solution

1

Identify key geometric properties of the kite's diagonals.

The diagonals

WY

and

XZ

are perpendicular at intersection point

M

, creating four right triangles inside the kite.

By definition, the diagonals of a kite are perpendicular to each other.

2

Calculate the lengths of the upper pair of equal sides (

WX

and

WZ

).

WX = \sqrt{WM^2 + XM^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ cm}

. Since

WZ = WX

,

WZ = 10\text{ cm}

.

Apply the Pythagorean theorem to right triangle

WMX

with legs of length 6 cm and 8 cm.

3

Calculate the lengths of the lower pair of equal sides (

YX

and

YZ

).

YX = \sqrt{MY^2 + XM^2} = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17\text{ cm}

. Since

YZ = YX

,

YZ = 17\text{ cm}

.

Apply the Pythagorean theorem to right triangle

YMX

with legs of length 15 cm and 8 cm.

4

Compute the total perimeter of kite

WXYZ

.

\text{Perimeter} = WX + WZ + YX + YZ = 10 + 10 + 17 + 17 = 54\text{ cm}

.

Sum the lengths of all four exterior sides.

Key Concept

Properties of Kite Diagonals and Pythagorean Theorem

Estimated Time:1m 15s

Question 174Question →

A rectangular billboard frame $ABCD$ has a length $AB = 16$ feet and width $BC = 12$ feet. A straight diagonal support brace connects vertex $A$ to vertex $C$ . To add structural stability, a secondary beam is installed perpendicular to diagonal $AC$ , extending from vertex $B$ to meet $AC$ at point $P$ . What is the length, in feet, of segment $BP$ ?

9.6

10.0

14.0

19.2

20.0

Show answer & explanation

Answer: 9.6

Answer

9.6 feet

The hypotenuse

AC

of right triangle

ABC

equals

\sqrt{16^2 + 12^2} = 20

feet. Since the area of triangle

ABC

can be calculated either as

\frac{1}{2} \times 16 \times 12 = 96

or as

\frac{1}{2} \times 20 \times BP

, solving

10 \times BP = 96

gives

BP = 9.6

feet.

Step-by-Step Solution

1

Calculate the length of diagonal

AC

using the Pythagorean theorem.

AC = \sqrt{AB^2 + BC^2} = \sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20

feet.

Triangle

ABC

is a right triangle with right angle at

B

and hypotenuse

AC

.

2

Express the area of triangle

ABC

using the two legs.

\text{Area} = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 16 \times 12 = 96

square feet.

The area of a right triangle is half the product of its perpendicular legs.

3

Express the area using hypotenuse

AC

as the base and

BP

as the height, then solve for

BP

.

\text{Area} = \frac{1}{2} \times AC \times BP \implies 96 = \frac{1}{2} \times 20 \times BP \implies 10 \times BP = 96 \implies BP = 9.6

feet.

Segment

BP

is given as perpendicular to base

AC

.

Key Concept

Altitude to the Hypotenuse in a Right Triangle

Estimated Time:1m 15s

Question 175Question →

In parallelogram $ABCD$ , the diagonals $AC$ and $BD$ intersect at point $E$ . If $AE = 2x + 5$ , $EC = 5x - 7$ , $BE = 3y - 1$ , and $ED = y + 9$ , what is the length of diagonal $BD$ ?

Show answer & explanation

Answer: 28

Answer

The length of diagonal

BD

is 28.

Because the diagonals of a parallelogram bisect each other, the intersection point

E

divides diagonal

BD

into two equal segments (

BE = ED

). Equating the expressions gives

3y - 1 = y + 9

, which simplifies to

2y = 10

, so

y = 5

. Substituting

y = 5

into the expression for

BE

yields

BE = 14

. Since

BD

consists of

BE + ED

, the full length of diagonal

BD

is

14 + 14 = 28

.

Step-by-Step Solution

1

Apply the diagonal bisection property of parallelograms.

BE = ED

, so

3y - 1 = y + 9

.

The diagonals of any parallelogram bisect each other at their intersection point.

2

Solve the linear equation for

y

.

2y = 10 \implies y = 5

.

Subtract

y

from both sides and add 1 to both sides.

3

Calculate segment

BE

and total length

BD

.

BE = 3(5) - 1 = 14

, so

BD = 2 \times 14 = 28

.

Substitute

y = 5

into the segment length expression and double it for the full diagonal length.

Key Concept

Diagonals of a parallelogram bisect each other.

Question 176Question →

In parallelogram $PQRS$ , the measures of consecutive angles $\angle P$ and $\angle Q$ are $(4x + 15)^\circ$ and $(2x + 45)^\circ$ , respectively. What is the measure, in degrees, of $\angle R$ ?

20^\circ

75^\circ

85^\circ

95^\circ

105^\circ

Show answer & explanation

Answer: $95^\circ$

Answer

95^\circ

In any parallelogram, consecutive interior angles formed by parallel lines and a transversal are supplementary (sum to

180^\circ

). Adding the expressions for consecutive angles

\angle P

and

\angle Q

gives

(4x + 15) + (2x + 45) = 180^\circ

, which simplifies to

6x + 60 = 180^\circ

, giving

x = 20

. Substituting

x = 20

into the expression for

\angle P

yields

4(20) + 15 = 95^\circ

. Because opposite angles in a parallelogram are congruent,

\angle R

has the same measure as

\angle P

, which is

95^\circ

.

Step-by-Step Solution

1

Set up an equation using the consecutive angle property of parallelograms.

(4x + 15) + (2x + 45) = 180

Consecutive angles in any parallelogram are supplementary (their sum is

180^\circ

).

2

Solve the linear equation for

x

.

6x + 60 = 180 \implies 6x = 120 \implies x = 20

Combine like terms (

4x + 2x = 6x

and

15 + 45 = 60

) and isolate

x

.

3

Calculate the measure of

\angle P

.

m\angle P = 4(20) + 15 = 80 + 15 = 95^\circ$

Substitute

x = 20

into the expression given for

\angle P

.

4

Determine the measure of

\angle R

.

m\angle R = m\angle P = 95^\circ$

Opposite angles in a parallelogram are equal in measure.

Key Concept

Angle Relationships in Parallelograms

Estimated Time:1m 0s

Question 177Question →

In right triangle $ABC$ , the right angle is located at vertex $B$ , and the measure of angle $A$ is $30^\circ$ . The hypotenuse $AC$ has a length of 16 inches. Segment $BD$ is an altitude drawn from vertex $B$ perpendicular to hypotenuse $AC$ at point $D$ . What is the length, in inches, of segment $CD$ ?

Show answer & explanation

Answer: 4

Answer

The length of segment CD is 4 inches.

In right triangle ABC with angle A = 30°, the side opposite angle A (BC) is half the hypotenuse AC, so BC = 16 / 2 = 8 inches. Drawing altitude BD creates smaller right triangle BCD with right angle at D and angle C = 60°. This makes triangle BCD another 30°-60°-90° right triangle where segment BC = 8 inches is the hypotenuse. Segment CD lies opposite the 30° angle DBC, meaning CD is half of BC: 8 / 2 = 4 inches.

Step-by-Step Solution

1

Determine the length of leg BC in right triangle ABC

BC = 8 inches

In a 30°-60°-90° triangle, the length of the side opposite the 30° angle is equal to half the length of the hypotenuse. Since hypotenuse AC = 16 inches, BC = 16 / 2 = 8 inches.

2

Identify the angles of right triangle BCD

Angle C = 60°, Angle BDC = 90°, and Angle DBC = 30°

Since angle A = 30° in right triangle ABC, angle C must equal 90° - 30° = 60°. Altitude BD creates right angle BDC = 90°, leaving angle DBC = 180° - 90° - 60° = 30°.

3

Calculate the length of segment CD in 30°-60°-90° triangle BCD

CD = 4 inches

In triangle BCD, segment BC (8 inches) is the hypotenuse. Segment CD lies opposite the 30° angle DBC, so its length is half of the hypotenuse BC: 8 / 2 = 4 inches.

Key Concept

Altitude to Hypotenuse in Special 30°-60°-90° Right Triangles

Question 178Question →

Determine whether the following statement regarding quadrilateral properties is true or false: Any convex quadrilateral whose diagonals intersect at right angles must be a rhombus.

True

False

Show answer & explanation

Answer: False

Answer

The statement is false. Perpendicular diagonals alone are not sufficient to prove that a quadrilateral is a rhombus.

The statement is false because perpendicular diagonals are not a sufficient condition to classify a general quadrilateral as a rhombus. Other quadrilaterals, such as kites or general orthodiagonal quadrilaterals with unequal sides, also have diagonals that intersect at right angles.

Step-by-Step Solution

1

Recall the defining properties of a rhombus.

A rhombus is a parallelogram with four congruent sides. Its diagonals are perpendicular bisectors of each other.

To evaluate whether perpendicular diagonals alone guarantee a rhombus.

2

Analyze counterexamples of non-rhombus quadrilaterals with perpendicular diagonals.

A kite has diagonals that intersect at right angles, but only adjacent pairs of sides are congruent, not all four sides. Furthermore, a general quadrilateral can have perpendicular diagonals of arbitrary lengths that do not bisect each other, resulting in four sides of completely different lengths.

A single counterexample disproves a universal mathematical claim.

3

Determine the overall truth value of the statement.

Because perpendicular diagonals are a necessary property of rhombuses but not a sufficient condition for all quadrilaterals, the statement is false.

Concluding the analysis based on geometric counterexamples.

Key Concept

Necessary vs. Sufficient Conditions for Quadrilateral Classification

Question 179Question →

In right triangle $XYZ$ , $\angle Y = 90^\circ$ and $\angle X = 45^\circ$ . The hypotenuse $XZ$ has a length of $12\sqrt{2}$ units. Point $W$ lies on leg $XY$ such that $XW = 7$ units. What is the length, in units, of segment $ZW$ ?

13

\sqrt{119}

17

\sqrt{193}

\sqrt{313}

Show answer & explanation

Answer: $13$

Answer

The length of segment

ZW

is

13

units.

Because

\triangle XYZ

is an isosceles right triangle (

45^\circ-45^\circ-90^\circ

), its leg lengths

XY

and

YZ

are equal to the hypotenuse divided by

\sqrt{2}

, giving

12

. Segment

YW

is

12 - 7 = 5

. In right triangle

\triangle ZYW

, the hypotenuse

ZW = \sqrt{12^2 + 5^2} = 13

.

Step-by-Step Solution

1

Determine the leg lengths of

\triangle XYZ

using special right triangle properties.

XY = YZ = 12

In a

45^\circ-45^\circ-90^\circ

right triangle, the hypotenuse is

\text{leg} \cdot \sqrt{2}

. Given

XZ = 12\sqrt{2}

, each leg length is

12

.

2

Calculate the length of segment

YW

.

YW = 5

Since point

W

lies on leg

XY

and

XW = 7

,

YW = XY - XW = 12 - 7 = 5

.

3

Apply the Pythagorean Theorem to right triangle

\triangle ZYW

to solve for hypotenuse

ZW

.

ZW = 13

ZW = \sqrt{YZ^2 + YW^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13

.

Key Concept

Properties of

45^\circ-45^\circ-90^\circ

special right triangles and multi-step applications of the Pythagorean Theorem.

Estimated Time:1m 0s

Question 180Question →

In rhombus $JKLM$ , diagonals $JL$ and $KM$ intersect at point $P$ . If $JP = 2x + 3$ , $PL = 4x - 5$ , and $m\angle JKL = 60^\circ$ , what is the length of side $JK$ ?

8

11

22

11\sqrt{3}

44

Show answer & explanation

Answer: $22$

Answer

The length of side

JK

is

22

.

Since the diagonals of a rhombus bisect each other, segment

JP

equals segment

PL

. Equating

2x + 3 = 4x - 5

yields

x = 4

, which gives

JP = 11

. The diagonals of a rhombus are perpendicular and bisect the vertex angles, creating right triangle

\triangle JPK

with

m\angle JPK = 90^\circ

and

m\angle JKP = 30^\circ

. In a

30^\circ

-

60^\circ

-

90^\circ

right triangle, the hypotenuse is twice the side opposite the

30^\circ

angle. Because

JP = 11

is opposite the

30^\circ

angle, side

JK = 2 \times 11 = 22

. Alternatively, since

\triangle JKL

is an isosceles triangle with a

60^\circ

vertex angle, it is equilateral, making

JK = JL = 22

.

Step-by-Step Solution

1

Use the diagonal bisecting property of a rhombus to set up an algebraic equation.

2x + 3 = 4x - 5 \implies 2x = 8 \implies x = 4

The diagonals of a rhombus bisect each other, so

JP = PL

.

2

Calculate the length of the half-diagonal segment

JP

.

JP = 2(4) + 3 = 11

Substitute

x = 4

into the expression for

JP

.

3

Determine the angle measures in right triangle

\triangle JPK

.

m\angle JPK = 90^\circ

and

m\angle JKP = \frac{1}{2}(60^\circ) = 30^\circ

The diagonals of a rhombus are perpendicular to each other and bisect the vertex angles.

4

Apply the

30^\circ

-

60^\circ

-

90^\circ

right triangle ratio or sine function to find hypotenuse

JK

.

\sin(30^\circ) = \frac{JP}{JK} \implies \frac{1}{2} = \frac{11}{JK} \implies JK = 22$

In right triangle

\triangle JPK

,

JP

is opposite the

30^\circ

angle, so the hypotenuse

JK

is twice the length of

JP

.

Key Concept

Properties of Rhombus Diagonals and Special Right Triangles

Estimated Time:1m 30s

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